Integrand size = 21, antiderivative size = 115 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac {a \cos (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sin (c+d x))} \]
arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(3/2)/d-1/2*cos (d*x+c)/b/d/(a+b*sin(d*x+c))^2+1/2*a*cos(d*x+c)/b/(a^2-b^2)/d/(a+b*sin(d*x +c))
Time = 0.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {2 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {\cos (c+d x) (b+a \sin (c+d x))}{(a+b \sin (c+d x))^2}}{2 (a-b) (a+b) d} \]
((2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (C os[c + d*x]*(b + a*Sin[c + d*x]))/(a + b*Sin[c + d*x])^2)/(2*(a - b)*(a + b)*d)
Time = 0.47 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3172, 3042, 3233, 27, 3042, 3139, 1083, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^2}{(a+b \sin (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3172 |
| \(\displaystyle -\frac {\int \frac {\sin (c+d x)}{(a+b \sin (c+d x))^2}dx}{2 b}-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\sin (c+d x)}{(a+b \sin (c+d x))^2}dx}{2 b}-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle -\frac {-\frac {\int \frac {b}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {a \cos (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 b}-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {b \int \frac {1}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {a \cos (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 b}-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {b \int \frac {1}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {a \cos (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 b}-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle -\frac {-\frac {2 b \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {a \cos (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 b}-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {\frac {4 b \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{d \left (a^2-b^2\right )}-\frac {a \cos (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 b}-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {-\frac {2 b \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac {a \cos (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 b}-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
-1/2*Cos[c + d*x]/(b*d*(a + b*Sin[c + d*x])^2) - ((-2*b*ArcTan[(2*b + 2*a* Tan[(c + d*x)/2])/(2*Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d) - (a*Cos[c + d*x])/((a^2 - b^2)*d*(a + b*Sin[c + d*x])))/(2*b)
3.5.58.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + 1))), x] + Simp[g^2*((p - 1)/(b*(m + 1))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; Fre eQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && I ntegersQ[2*m, 2*p]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Time = 1.04 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.75
| method | result | size |
| derivativedivides | \(\frac {\frac {-\frac {\left (a^{2}-2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \left (a^{2}-b^{2}\right )}+\frac {\left (a^{2}+2 b^{2}\right ) b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{2}-b^{2}\right ) a^{2}}+\frac {\left (a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (a^{2}-b^{2}\right )}+\frac {2 b}{2 a^{2}-2 b^{2}}}{{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {\arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{d}\) | \(201\) |
| default | \(\frac {\frac {-\frac {\left (a^{2}-2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \left (a^{2}-b^{2}\right )}+\frac {\left (a^{2}+2 b^{2}\right ) b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{2}-b^{2}\right ) a^{2}}+\frac {\left (a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (a^{2}-b^{2}\right )}+\frac {2 b}{2 a^{2}-2 b^{2}}}{{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {\arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{d}\) | \(201\) |
| risch | \(-\frac {i \left (-2 i a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+i b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+2 i a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+i b^{3} {\mathrm e}^{i \left (d x +c \right )}+2 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-a \,b^{2}\right )}{\left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+i b \right )^{2} \left (a^{2}-b^{2}\right ) d \,b^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\) | \(307\) |
1/d*(2*(-1/2*(a^2-2*b^2)/a/(a^2-b^2)*tan(1/2*d*x+1/2*c)^3+1/2*(a^2+2*b^2)* b/(a^2-b^2)/a^2*tan(1/2*d*x+1/2*c)^2+1/2*(a^2+2*b^2)/a/(a^2-b^2)*tan(1/2*d *x+1/2*c)+1/2*b/(a^2-b^2))/(a*tan(1/2*d*x+1/2*c)^2+2*b*tan(1/2*d*x+1/2*c)+ a)^2+1/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^( 1/2)))
Time = 0.33 (sec) , antiderivative size = 501, normalized size of antiderivative = 4.36 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\left [-\frac {2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{4 \, {\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d \sin \left (d x + c\right ) - {\left (a^{6} - a^{4} b^{2} - a^{2} b^{4} + b^{6}\right )} d\right )}}, -\frac {{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d \sin \left (d x + c\right ) - {\left (a^{6} - a^{4} b^{2} - a^{2} b^{4} + b^{6}\right )} d\right )}}\right ] \]
[-1/4*(2*(a^3 - a*b^2)*cos(d*x + c)*sin(d*x + c) - (b^2*cos(d*x + c)^2 - 2 *a*b*sin(d*x + c) - a^2 - b^2)*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d* x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) + 2*(a^2*b - b^3)*cos(d*x + c))/((a^4*b^2 - 2*a^2*b^4 + b^6)*d*cos(d*x + c)^2 - 2*(a^5*b - 2*a^3*b^3 + a*b^5)*d*sin(d*x + c) - (a^ 6 - a^4*b^2 - a^2*b^4 + b^6)*d), -1/2*((a^3 - a*b^2)*cos(d*x + c)*sin(d*x + c) + (b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)*sqrt(a^2 - b^ 2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + (a^2*b - b^3)*cos(d*x + c))/((a^4*b^2 - 2*a^2*b^4 + b^6)*d*cos(d*x + c)^2 - 2*(a^5 *b - 2*a^3*b^3 + a*b^5)*d*sin(d*x + c) - (a^6 - a^4*b^2 - a^2*b^4 + b^6)*d )]
Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.54 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.80 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} - \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2} b}{{\left (a^{4} - a^{2} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}}}{d} \]
((pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(a^2 - b^2)^(3/2) - (a^3*tan(1/2*d*x + 1/2*c)^3 - 2*a*b^2*tan(1/2*d*x + 1/2*c)^3 - a^2*b*tan(1/2*d*x + 1/2*c)^2 - 2*b^3*tan( 1/2*d*x + 1/2*c)^2 - a^3*tan(1/2*d*x + 1/2*c) - 2*a*b^2*tan(1/2*d*x + 1/2* c) - a^2*b)/((a^4 - a^2*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2))/d
Time = 6.93 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.45 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {b}{a^2-b^2}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^2-2\,b^2\right )}{a\,\left (a^2-b^2\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+2\,b^2\right )}{a\,\left (a^2-b^2\right )}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2+2\,b^2\right )}{a^2\,\left (a^2-b^2\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2+4\,b^2\right )+a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+a^2+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {\mathrm {atan}\left (\left (a^2-b^2\right )\,\left (\frac {a^2\,b-b^3}{{\left (a+b\right )}^{3/2}\,\left (a^2-b^2\right )\,{\left (a-b\right )}^{3/2}}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \]
(b/(a^2 - b^2) - (tan(c/2 + (d*x)/2)^3*(a^2 - 2*b^2))/(a*(a^2 - b^2)) + (t an(c/2 + (d*x)/2)*(a^2 + 2*b^2))/(a*(a^2 - b^2)) + (b*tan(c/2 + (d*x)/2)^2 *(a^2 + 2*b^2))/(a^2*(a^2 - b^2)))/(d*(tan(c/2 + (d*x)/2)^2*(2*a^2 + 4*b^2 ) + a^2*tan(c/2 + (d*x)/2)^4 + a^2 + 4*a*b*tan(c/2 + (d*x)/2)^3 + 4*a*b*ta n(c/2 + (d*x)/2))) + atan((a^2 - b^2)*((a^2*b - b^3)/((a + b)^(3/2)*(a^2 - b^2)*(a - b)^(3/2)) + (a*tan(c/2 + (d*x)/2))/((a + b)^(3/2)*(a - b)^(3/2) )))/(d*(a + b)^(3/2)*(a - b)^(3/2))